Question

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m/s. At t = 0, an element of the string at x = 0 has a transverse position of 2.00 cm and is traveling downward with a speed of 2.00 m/s.
(a) What is the amplitude of the wave?
(b) What is the initial phase angle?
(c) What is the maximum transverse speed of an element of the string?
(d) Write the wave equation for the wave.

Answer 1

Answer:

a) A =0.021525m

b) $\phi=0.37869rad$

c) $v_{max}=5.4098\frac{m}{s}$

d)$y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

Explanation:

1) Notation

A= Amplitude

v= velocity

$\lambda$= wavelength

k= wave number

$\omega$= angular frequency

f= frequency

2) Part a and b

The equation of movement for a transverse sinusoidal wave is gyben by (1)

$y(t)=Acos(kx+ \omega t +\phi)$   (1)

At x=0 ,t=0 we have that:

$0.02=Acos(\phi)$

The velocity would be the derivate of the position, so taking the derivate of (1) respect to t we got (2)

$v(t)=-\omega Asin(kx+ \omega t+\phi)$   (2)

And replacing the conditions at x=0, t=0 we got

$-2\frac{m}{s}=-\omega Asin(\phi)$  

Now we can find the angular frequency with equation (3)

$\omega =\frac{2\pi}{T}$   (3)

Replacing the values obtained we got:

$\omega =\frac{2\pi}{0.025s}=80\pi \frac{rad}{s}$  

From equation (1) we have:

$Acos(\phi)=0.02$   (a)

$-2=-80\pi Asin(\phi)$   (b)

So from condition (b) we have:

$Asin(\phi)=\frac{1}{40\pi}$   (c)

If we divide condition (c) by condition (a) we got:

$\frac{Asin(\phi)}{Acos(\phi)}=tan(\phi)=\frac{1}{0.02x40\pi}=\frac{1}{0.8\pi}=0.39789$

If we solve for $\phi$ we got:

$\phi =tan^{-1}(0.39789)=0.37869$

And now since we have $\phi$ we can find A from equation (a)

$Acos(0.37869)=0.02$

So then Solving for A we got $A=\frac{0.02}{cos(0.37869)}=0.021525$

3) Part c

From equation (2) we can see that the maximum speed occurs when $sin(\omega t+\phi)=1$, so on this case we have:

$v_{max}=\omega A=80\pi \frac{rad}{s}x0.021525m=5.4098\frac{m}{s}$

4) Part d

On this case we need an equation like (1), and we have everything except the wave number, and we can obtain this from the following expression:

$v=\lambda f=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}$   (4)

And solving for k from equation (4) we got

$k=\frac{\omega}{v}=\frac{80\pi \frac{rad}{s}}{30\frac{m}{s}}=\frac{8\pi}{3}m^{-1}}$

And with the k number we have everythin in order to create the wave function, given by:

$y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$